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Find the greatest common divisor of $6x^2 +22$ and $2x+19$ in $\Bbb Z_{23}[x]$.

Computing I get that $$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{2 x}+19}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrr}{3 x}&{+6}&\phantom{+22}\end{array}\\{2 x}+19&\phantom{-}\enclose{longdiv}{\begin{array}{ccc}{6 x^{2}}&+0 x&+22\end{array}}&\\\phantom{\color{Magenta}{2 x}+19}&\begin{array}{rrr}-\phantom{6 x^{2}}&\phantom{+0 x}&\phantom{+22}\\\phantom{\enclose{longdiv}{}}6 x^{2}&+11 x\\\hline\phantom{\enclose{longdiv}{}}&{- 11 x}&+22\\&-\phantom{- 57 x}&\phantom{+22}\\\phantom{\enclose{longdiv}{}}&-11 x&+22\\\hline\phantom{\enclose{longdiv}{}}&&0\end{array}&\begin{array}{c}\phantom{6 x^{2}+0 x+22}\\\\\\\phantom{- 57 x+22}\\\color{Green}\\\phantom{\frac{1127}{2}}\end{array}\end{array}$$

And by the extended euclidean algorithm the last nonzero remainder is the gcd. However here I'm not getting a remainder at all so what is the gcd in this case?

The actual solution for the question is that the gcd should be $x+21$, but where can I derive this?

Cameron Williams
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Jacob Sanders
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    The gcd is the last nonzero remainder, here $-11x+22,,$ i.e. your computation implies that $\gcd(6x^2+22,2x+19) = \gcd(-11x+22,0) = -11x+22 = -11(x-2) = \color{#c00}{-11}(x+21),$ which unit-normalized monic is $,x+21,$ after cancelling the unit $\color{#c00}{-11}$ (cf. linked dupe). – Bill Dubuque Mar 30 '22 at 20:11
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    Or simply $2x+19=2x-4=\color{red}{2}(x-2)=\color{red}{2}(x+21)$ – zwim Mar 30 '22 at 20:16
  • Note: if you are doing complete (vs. single-term) divisions at each step in the Euclidean algorithm then - as zwim said - the last nonzero remainder is $2x+19 = 2x-4$ which unit-normalizes to $x-2 = x+21$ after cancelling $2$. Else it may be $-11x+22$ as in my first comment. If you are not familiar with optimizations/tweaks of the Euclidean algorithm it's probably best to go with the standard complete division steps used in most introductory expositions. – Bill Dubuque Mar 30 '22 at 20:35
  • @BillDubuque Thanks for the comment. Is there a way to get to the form $$a(x)(6x^2+22) + b(x)(2x+19) = -11x +22$$ from here somehow? I don't know how to find $a(x)$ and $b(x)$. – Jacob Sanders Mar 30 '22 at 21:45
  • Here we have the easy case $,\gcd(a,ab) = a,$ with Bezout equation $a = 1\cdot a + 0\cdot ab.,$ Generally it's easiest to use this forward form of the extended Euclidean algorithm. – Bill Dubuque Mar 30 '22 at 21:59

1 Answers1

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If $2+19$ divides $6^2+22$ $2+19$ is the common denominator. Find the common denominator of 21 and 7 you would know immediately.

trula
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  • What do you mean by finding the common denominator of $21$ and $7$? $7$ is a prime so the common denominator is either $1$ or $7$? And where did this $7$ come from? – Jacob Sanders Mar 30 '22 at 20:08